2. Use the table in Question 1 to solve \( x \) : \( 2.1 \quad 2^{x}=4 \) 2.2 \( 2.4 \quad 2^{x}=32 \) 2.5 2.7 \( 3^{x}=\frac{1}{8} \)

Let’s solve these equations one by one! For \( 2.1 \), \( 2^x = 4 \). Since \( 4 \) can be expressed as \( 2^2 \), we have \( 2^x = 2^2 \). Therefore, \( x = 2 \). For \( 2.4 \), \( 2^x = 32 \). Notice that \( 32 \) can be rewritten as \( 2^5 \). So, \( 2^x = 2^5 \) leads us to conclude that \( x = 5 \). Now for \( 2.7 \), \( 3^x = \frac{1}{8} \). To solve for \( x \), it's useful to express \( \frac{1}{8} \) in terms of base 3. However, we can actually take the logarithm of both sides: \( x = \log_3\left(\frac{1}{8}\right) = \log_3(8^{-1}) = -\log_3(8) \). Since \( 8 \) is \( 2^3 \), we can use \( \log_3(2) \) to further approximate this value if needed, but the exact numeric form is sufficient for now!